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PHYSICS, M2 EQ-Bank 7

A 75 kg cyclist rides up a hill with a vertical height of 45 m. The cyclist maintains a constant speed of 3.0 m/s along a path that is inclined at 8.0° to the horizontal.
 

  1. Calculate the power required to overcome gravity during this climb.   (2 marks)
  1. If the cyclist's muscles operate at an efficiency of 25% and the bicycle drivetrain has an efficiency of 95%, calculate the total power input required from the cyclist.   (2 marks)
Show Answers Only

a.    \(306.9\ \text{W}\)

b.    \(1.292\ \text{kW}\)

Show Worked Solution

a.    Find force of the rider down the slope of plane due to gravity:

   \(F=mg\ \sin \theta = 75 \times 9.8 \times \sin 8^{\circ} = 102.3\ \text{N}\)

Find power required to overcome this force using \(P= Fv\):

   \(\text{Power}\ = F \times v = 102.5 \times 3.0 = 306.9\ \text{W}\)

 
b.    Total power input required:

  • Power output to overcome gravity \(= 306.9\ \text{W}\)
  • Power after drivetrain efficiency \(= \dfrac{306.9}{0.95} = 323.05\ \text{W}\)
  • Total power input (incl. muscle efficiency) \(= \dfrac{323.05}{0.25} = 1292.2\ \text{W} = 1.292\ \text{kW}\)

PHYSICS, M2 EQ-Bank 6

A crate, initially at rest, is pushed across a horizontal, frictionless floor by a constant force of 60 N for 8 seconds. During this time, the crate accelerates at a rate of 3.0 ms\(^{-2}\) (assume the force is parallel to the direction of motion).

What is the average power exerted on the crate at the end during the 8 seconds?   (2 marks)

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\(720\ \text{W}\)

Show Worked Solution
  • Find the distance that the crate travels:
  •    \(s=ut +\dfrac{1}{2}at^2=0 \times 3 +\dfrac{1}{2} \times 3 \times 8^2=96\ \text{m}\)
  • Work Done on crate \(=Fs = 60 \times 96 = 5760\ \text{J}\)
  • Find average power exerted on the crate:
  •    \(P_{\text{avg}} = \dfrac{\text{Work Done}}{t}= \dfrac{5760}{8} = 720\ \text{W}\).

PHYSICS, M2 EQ-Bank 9 MC

A construction hoist uses an electric motor with a power rating of 6.0 kW, which operates at 75% efficiency when lifting materials vertically.

What is the maximum speed at which the hoist can raise a 240 kg load?

  1. \(3.1\ \text{ms}^{-1}\)
  2. \(2.0\ \text{ms}^{-1}\)
  3. \(1.5\ \text{ms}^{-1}\)
  4. \(1.9\ \text{ms}^{-1}\)
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\(D\)

Show Worked Solution
  • Power lifting mass vertically \(= 0.75 \times 6 = 4.5\ \text{kW} = 4500\ \text{W}\).
\(P\) \(=F_{\parallel}v\)  
\(v\) \(=\dfrac{P}{F_{\parallel}}=\dfrac{4500}{240 \times 9.8}=1.9\ \text{ms}^{-1}\)  

 
\(\Rightarrow D\)